Question

Find the probability that the sum of the $40$ values is greater than$7,500$.

Short answer

The probability that the sum of the $40$ values is greater than $7500$is$\mathrm{P}(X\ge 7500)=0.0089$.

Step-by-step solution

Given Information

A mean (${\mu }_x$)is$180$and a standard deviation (${\sigma }_x$) is $20$and sample size $\left(n\right)$is$40$.

Explanation

To find the probability that the sum of the $40$values is less than that $7500$:

$\sum X~N\left(n{\mu }_x,\sqrt{n}{\sigma }_x\right)$

$\sum X~N\left(\right(40\left)\right(180),(\sqrt{40}\left)\right(20\left)\right)$

$\mathrm{P}(X\ge 7500)=\mathrm{P}\left(Z\ge \frac{X-n{\mu }_x}{\sqrt{n}{\sigma }_x}\right)=\mathrm{P}\left(Z\ge \frac{7500-7200}{126.4911}\right)=\mathrm{P}(Z\ge 2.3717)$

$\mathrm{P}(X\ge 7500)=0.0089$